A sample containing 2.20 mol of Ne gas has an initial volume of 8.00 L . What is

Question

A sample containing 2.20 mol of Ne gas has an initial volume of 8.00 L . What is the final volume, in liters, when the following changes occur in the quantity of the gas at constant pressure and temperature?

Part A

Part complete

A leak allows one-half of Ne atoms to escape.

Express your answer with the appropriate units.

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Part B

Part complete

A sample of 4.00 mol of Ne is added to the 2.20 mol of Ne gas in the container.

Express your answer with the appropriate units.

L

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Part C

Part complete

A sample of 21.0 g of Ne is added to the 2.20 mol of Ne gas in the container.

Express your answer with the appropriate units.

L

   L

A sample containing 2.20 mol of Ne gas has an initial volume of 8.00 L . What is the final volume, in liters, when the following changes occur in the quantity of the gas at constant pressure and temperature?

Part A

Part complete

A leak allows one-half of Ne atoms to escape.

Express your answer with the appropriate units.

SubmitPrevious AnswersRequest Answer

Part B

Part complete

A sample of 4.00 mol of Ne is added to the 2.20 mol of Ne gas in the container.

Express your answer with the appropriate units.

V =

L

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Part C

Part complete

A sample of 21.0 g of Ne is added to the 2.20 mol of Ne gas in the container.

Express your answer with the appropriate units.

V =

L

V =

   L

Explanation / Answer

A)
Given:
Vi = 8.00 L
ni = 2.20 mol
nf = 1.10 mol

use:
Vi/ni = Vf/nf
8.00 L / 2.20 mol = Vf / 1.10 mol
Vf = 4.00 L
Answer: 4.00 L

B)
Given:
Vi = 8.00 L
ni = 2.20 mol
nf = 6.20 mol

use:
Vi/ni = Vf/nf
8.00 L / 2.20 mol = Vf / 6.20 mol
Vf = 22.5 L
Answer: 22.5 L

C)

Molar mass of Ne = 20.18 g/mol


mass(Ne)= 21.0 g

use:
number of mol of Ne,
n = mass of Ne/molar mass of Ne
=(21.0 g)/(20.18 g/mol)
= 1.04 mol

Given:
Vi = 8.00 L
ni = 2.20 mol
nf = 3.24 mol

use:
Vi/ni = Vf/nf
8.00 L / 2.20 mol = Vf / 3.24 mol
Vf = 11.8 L
Answer: 11.8 L

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